## Download PDF of ICSE Class 10 Similarity Previous Years Questions

## ICSE Class 10 Similarity Previous Years Questions+Solution

**1. In the given figure, AC ∥ DE ∥ BF. If AC = 24 cm, EG = 8 cm, GB = 16 cm, BF = 30 cm,**

**(a) prove ΔGED ~ ΔGBF**

**(b) find DE.**

**(c) DB : AB**

**Solution:** **(b) 15 cm (c) 5 : 8**

**Step-by-step Explanation:**

(a) In ΔGED and ΔGBF,

∠EGD = ∠BGF (vertically opposite angles)

∠GED = ∠GBF (alternate interior angles)

∴ ΔGED ~ ΔGBF (by A-A axiom of similarity)

(b) as ΔGED ~ ΔGBF

∴ GE/BG = DE/BF = DG/GF

GE/BG = DE/BF

8/16 = DE/30

DE = 8×30/16

DE = 15 cm

(c) In ΔDBE and ΔABC,

∠B = ∠B (common)

∠EDB = ∠CAB (corresponding angles)

∴ ΔDBE ~ ΔABC (by A-A condition of similarity)

∴ DB/AB = BE/BC = DE/AC

DB/AB = DE/AC

DB/AB = 15/24 = 5/8

DB : AB = 5 : 8

**2. In the given figure, ∠BAP = ∠DCP = 70°, PC = 6 cm and CA = 4 cm, then PD : DB is**

**(a) 5 : 3**

**(b) 3 : 5**

**(c) 3 : 2**

**(d) 2 : 3 [2023]**

**Solution: (c) 3 : 2**

**Step-by-step Explanation:**

In ΔPCD and ΔPAB,

∠PCD = ∠BAP = 70**°,** (given)

∠DPC = ∠BPA (common)

∴ ΔPCD ~ ΔPAB (by A-A condition of similarity)

∴ PC/PA = PD/PB = CD/AB

PC/PA = PD/PB

6/6+4 = 6/10 = 3/5 = PD/PB

As PD : PB = 3 : 5, therefore PD : DB = 3 : (5-3) = 3 : 2

option (c) is correct.

**ICSE Class 10 Maths PYQs chapter-wise**

**3. In the given figure, PQ is parallel to TR, then by using condition of similarity:**

**(a) PQ/RT = OP/OT = OQ/OR**

**(b) PQ/RT = OP/OR = OQ/OT**

**(c) PQ/RT = OR/OP = OQ/OT**

**(d) PQ/RT = OP/OR = OT/OQ**

**[2021 SEMESTER-1]**

**Solution:** **(b) PQ/RT = OP/OR = OQ/OT**

**Step-by-step Explanation:**

In ΔPOQ and ΔROT,

∠POQ = ∠ROT (vertically opposite angles)

∠OPQ = ∠ORT (alternate interior angles)

∴ ΔPOQ ~ ΔROT (by A-A condition of similarity)

so, PQ/RT = OP/OR = OQ/OT

**Video solution of ICSE Class 10 Similarity Previous Years Questions**

**4. In the given figure ABCD is a trapezium in which DC is parallel to AB. AB = 16 cm and DC = 8 cm. OD = 5 cm, OB = (y + 3) cm, OA = 11 cm and OC = (x – 1) cm. Using the given information answer the following questions. [2021 Semester-I]**

**(i.) From the given figure name the pair of similar triangles: (a) ∆OAB, ∆OBC (b) ∆COD, ∆AOB (c) ∆ADB, ∆ACB (d) ∆COD, ∆COB**

**(ii.) The corresponding proportional sides with respect to the pair of similar triangles obtained in (i): (a) CD/AB = OC/OA = OD/OB (b) AD/BC = OC/OA = OD/OB **

** (c) AD/BC = BD/AC = AB/DC (d)OD/OB = CD/CB = OC/OA**

**(iii.) The ratio of the sides of the pair of similar triangles is: (a) 1 : 3 (b) 1 : 2 (c) 2 : 3 (d) 3 : 1**

**(iv.) Using the ratio of sides of the pair of similar triangles values of x and y are respectively: (a) x = 4.6, y = 7 (b) x = 7, y = 7 (c) x = 6.5, y = 7 (d) x = 6.5, y = 2**

**Solution: (i) (b) (ii) (a) (iii) (b) (iv) (c) **

**Step-by-step Explanation:**

In ΔCOD and ΔAOB,

∠COD = ∠AOB (vertically opposite angles)

∠OCD = ∠OAB (alternate interior angles)

∠POQ = ∠ROT (vertically opposite angles)

∴ ΔCOD ~ ΔAOB (by A-A condition of similarity)

Option (b) is correct.

(ii) Therefore, CD/AB = OC/OA = OD/OB

option (a) is correct.

(iii) CD/AB = 8/16 = 1 : 2

option (b) is correct.

(iv) We know, CD/AB = OC/OA = OD/OB

CD/AB = OC/OA

1/2 = x-1/11

2(x-1) = 11

2x-2 = 11

2x = 11+2

**x = 13/2 = 6.5 cm**

Now, CD/AB = OD/OB

1/2 = 5/y+3

y+3 = 10

**y = 7**

option (c) is correct.

**5. In the given figure, AB = 24 cm, AC = 18 cm, DE = 12cm, DF = 9 cm and ∠BAC = ∠EDF. Then ΔABC ~ ΔDEF by the condition:(a) AAA(b) SAS(c) SSS(d) AAS [2021 Semester-1]**

**Solution: (b)**

**Step-by-step Explanation:**

In ΔABC and ΔDEF,

AC/DF = 18/9 =2/1

∠BAC = ∠EDF (given)

AB/DE = 24/12 = 2/1

∴ ΔABC ~ ΔDEF (by S-A-S condition of similarity)

option (b) is correct.

**6. In the given figure, ∠PQR = ∠PST = 90°, PQ = 5 cm and PS = 2 cm.**

**(i) Prove that △PQR ∼ △PST.(ii) Find Area of △PQR: Area of quadrilateral SRQT. [2019]**

**Solution: (ii) 25 : 21**

**Step-by-step Explanation:**

(i) In ΔPQR and △PST,

∠P is common.

∠PQR= ∠PST = 90**°** (given)

∴ ΔPQR ~ ΔPST (A-A condition of similarity)

(ii) ∴ PQ/PS = QR/ST = PR/PT

As ΔPQR ~ ΔPST, therefore

Area of ΔPQR / Area of ΔPST =(PQ/PS)^{2} =(5/2)^{2} =25 : 4

∴ Area of ΔPQR / Area of quadrilateral SRQT = Area of ΔPQR / Area of ΔPQR-area of ΔPST

=25/25-4 = 25/21 = 25 : 21

**7. In ∆ PQR, MN is parallel to QR and PM/MQ = 2/3 [3](i) Find MN / QR(ii) Prove that ∆OMN and ∆ORQ are similar.(iii) Find, Area of ∆OMN: Area of ∆ORQ [2018]**

**Solution: (i) 2/5 (iii) 4 : 25 **

**Step-by-step Explanation:**

(i) In ΔPMN and △PQR,

∠P is common.

∠PMN= ∠PQR (corresponding angles)

∴ ΔPMN ~ ΔPQR (A-A condition of similarity)

We know, PM/MQ = 2/3

∴ PM/PQ = PM/PM + MQ = 2/2+3 = 2/5

∴ PM/PQ = MN/QR = PN/PR = 2/5

∴ MN/QR = 2/5

(ii) In ΔOMN and △ORQ,

∠MON = ∠ROQ (vertically opposite angles)

∠OMN= ∠ORQ (alternate interior angles)

∴ ΔOMN ~ ΔORQ (A-A condition of similarity)

(iii) Area of ΔOMN / Area of ΔORQ =(MN/QR)^{2} =(2/5)^{2} = 4 : 25

**8. PQR is a triangle. S is a point on the side QR of ∆PQR such that ∠PSR = ∠QPR. Given QP = 8 cm, PR = 6 cm and SR = 3 cm.**

**(i) Prove ∆PQR ~ ∆SPR(ii) Find the length of QR and PS(iii) area of ΔPQR/ area of ΔSPR [2017]**

**Solution: (ii) OR=12cm, PS = 4cm (iii) 4 : 1**

**Step-by-step Explanation:**

(i) In ΔPQR and △SPR,

∠R is common.

∠PSR = ∠QPR (given)

∴ ΔPQR ~ ΔSPR (A-A condition of similarity)

(ii) ∴ PQ/PS = QR/PR = PR/SR

PQ/PS = PR/SR

8/PS = 6/3

PS = 4 cm

QR/PR = PR/SR

QR/6 = 6/3

QR = 12 cm

(iii) Area of ΔPQR / Area of ΔSPR =(PR/SR)^{2} =(6/3)^{2} = 4 : 1

**9. In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T. **

**(i) Prove ∆ TPS ~ ∆ TRQ.(ii) Find SP if TP= 18cm, RQ= 4cm and TR= 6cm.(iii) Find area of quadrilateral PQRS if area of ∆ PTS = 27 cm**

^{2}[2016]

**Solution: (ii) 12 cm (iii) 24 cm ^{2} **

**Step-by-step Explanation:**

(i) In ΔTPS and △TRQ,

∠T is common.

∠TPS = ∠TRQ (exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

∴ ΔTPS ~ ΔTRQ (A-A condition of similarity)

(ii) ∴ TP/TR = SP/RQ = TS/TQ

TP/TR = SP/RQ

18/6 = SP/4

SP = 12 cm

(iii) As ΔTPS ~ ΔTRQ, therefore

Area of ΔTPS / Area of ΔTRQ =(TP/TR)^{2} =(18/6)^{2} = 9/1

27/Area of ΔTRQ = 9/1

Area of ΔTRQ = 3 cm^{2}

∴ area of quadrilateral PQRS = 27 – 3 = 24 cm^{2}

**10. ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC.**

**Prove that:(i) ∆ADE ~ ∆ACB.(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.(iii) Find, area of ∆ ADE: area of quadrilateral BCED. [2015]**

**Solution: (ii) DE = 1 2/3 cm, AD = 4 1/3 cm (iii) 1 : 8 **

**Step-by-step Explanation:**

(i) In ΔADE and △ACB,

∠A is common.

∠AED = ∠ABC = 90**°** (given)

∴ ΔADE ~ ΔACB (A-A condition of similarity)

(ii) ∴ AD/AC = DE/BC = AE/AB

In right-angled ΔACB, by pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

13^{2} = AB^{2} + 5^{2}

AB = √169 – 25 = √144 = 12 cm

Now, AD/AC = AE/AB

AD/13 = 4/12

AD = 13/3 = 4 1/3 cm

DE/BC = AE/AB

DE/5 = 4/12

DE = 5/3 = 1 2/3 cm

(iii) As ΔADE ~ ΔACB, therefore

Area of ΔADE / Area of ΔACB =(AE/AB)^{2} =(4/12)^{2} = 1 : 9

∴ Area of ΔADE / Area of quadrilateral BCED = Area of ΔADE / Area of ΔABC – area of ΔADE

= 1/9-1= 1/8= 1 : 8

**11. In ∆ABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm, AD = 5 cm.**

(i) Prove that ∆ACD is similar to ∆BCA.

(ii) Find BC and CD

(iii) Find area of ∆ACD : area of ∆ABC.

**Solution: (ii) BC = 6.4 cm, CD = 2.5 cm (iii) 25 : 64 **

**Step-by-step Explanation:**

(i) In ΔACD and △BCA,

∠C is common.

∠DAC = ∠ABC (given)

∴ ΔACD ~ ΔBCA (A-A condition of similarity)

(ii) ∴ AC/BC = CD/AC = AD/AB

CD/AC = AD/AB

CD/4 = 5/8

CD = 5/2 = 2.5 cm

AC/BC = AD/AB

4/BC = 5/8

BC = 32/5 = 6.4 cm

(iii) Area of ΔACD / Area of ΔABC =(AD/AB)^{2} =(5/8)^{2} = 25 : 64

**12. In the given figure, AB and DE are perpendiculars to BC.**

**(i) Prove that ΔABC ~ ΔDEC**

**(ii) If AB = 6 cm, DE = 4 cm and AC = 15 cm. Calculate CD.**

**(iii) Find the ratio of the area of ΔABC : area of ΔDEC**

**Solution: (ii) CD = 10 cm (iii) 9 : 4**

**Step-by-step Explanation:**

(i) In ΔABC and △DEC,

∠C is common.

∠ABC = ∠DEC = 90**°** (given)

∴ ΔABC ~ ΔDEC (A-A condition of similarity)

(ii) ∴ AB/DE = BC/EC = AC/CD

AB/DE = AC/CD

6/4 = 15/CD

CD = 10 cm

(iii) Area of ΔABC / Area of ΔDEC =(AB/DE)^{2} =(6/4)^{2} = 9 : 4

**13. In the given figure Δ ABC and Δ AMP are right angled at B and M respectively.Given, AB = 10 cm, AP = 15 cm and PM = 12 cm**

**(i) Prove that Δ ABC ~ Δ AMP(ii) Find AC and BC. [2012]**

**Solution: (ii) AC= 16 2/3 cm BC = 13 1/3 cm **

**Step-by-step Explanation:**

(i) In ΔABC and △AMP,

∠A is common.

∠ABC = ∠AMP = 90**°** (given)

∴ ΔABC ~ ΔAMP (A-A condition of similarity)

(ii) In right-angled △AMP, by pythagoras theorem,

AP^{2} = AM^{2} + PM^{2}

15^{2} = AM^{2} + 12^{2}

AM = √225 – 144 = √81 = 9 cm

now, as ΔABC ~ ΔAMP

∴ AB/AM = BC/PM = AC/AP

AB/AM = BC/PM

10/9 = BC/12

BC =40/3 = 13 1/3 cm

AB/AM = AC/AP

10/9 = AC/15

AC = 50/3 = 16 2/3 cm

**14. In the adjoining figure ABC is a right-angled triangle with ∠BAC = 90°,**

(i) Prove Δ ADB ~ Δ CDA

(ii) If BD = 18 cm and CD = 8 cm, find AD.

(iii) Find the ratio of area of Δ ADB is to area of Δ CDA. [2011]

**Solution: (ii) AD = 12 cm (iii) 9 : 4 **

**Step-by-step Explanation:**

(i) In ΔADB, ∠ABD = 180**°**-(∠ADB + ∠DAB) = 90**°**– ∠DAB

In ΔCDA, ∠CAD = 90**°**– ∠DAB

Hence, ∠CAD = ∠ABD

In ΔADB and △CDA,

∠ADB = ∠CDA = 90**°** (given)

∠CAD = ∠ABD (proved above)

∴ ΔADB ~ ΔCDA (A-A condition of similarity)

(ii) ∴ AD/CD = BD/AD = AB/AC

AD/CD = BD/AD

AD/8 =18/AD

AD^{2} =18×8

AD = √144 = 12 cm

(iii) Area of ΔADB / Area of ΔCDA =(AD/CD)^{2} =(12/8)^{2} = 9 : 4

**15. In the figure ABC is a triangle with ∠EDB = ∠ACB.**

Prove that Δ ABC ~ Δ EBD

If BE = 6 cm, EC = 4 cm BD = 5 cm and area of ΔBED = 9 cm^{2} ,

Calculate the (i) length of AB

(ii) area of Δ ABC [2010]

**Solution: (i) 12 cm (ii) 36 cm ^{2} **

**Step-by-step Explanation:**

In ΔABC and △EBD,

∠B is common.

∠ACB = ∠BDE (given)

∴ ΔABC ~ ΔEBD (A-A condition of similarity)

∴ AB/BE = BC/BD = AC/ED

(i) so, AB/BE = BC/BD

AB/6 = 6+4/5

AB/6 = 10/5

AB = 12 cm

(iii) Area of ΔABC / Area of ΔEBD =(AB/BE)^{2} =(12/6)^{2}

Area of ΔABC / 9 = 4/1

Hence, Area of ΔABC = 36 cm^{2}

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